Mathematical Models - Assignment Example

CFD ASSIGNMENT Mathematical models a) The governing equations of fire simulation are represented by the conservation equationsof mass, energy and momentum for fluids which obey Newton’s Law of Viscosity. This therefore implies that the flame is taken as a fluid whose shear stress is directly proportional to its rate of shear strain or velocity gradient. 1. Conservation of Mass This equation states that for any closed system, the mass is constant throughout. That is; the fluid generated in the system is equal to the mass flow in and out of the system together with the accompanying expansion or contraction of the system. In solving the above equation, the prevailing boundary conditions must be considered so as to ensure accuracy of the results is maintained. Due to the fact that this equation is held in a closed system, only the prevailing wall boundary conditions are of utmost importance and need to be considered for a complete simulation (Ferziger and Peric, 1999). 2. Conservation of Momentum () This equation states that the acceleration, convection and pressure gradient of a fire in motion is equal to the force of gravity acting on it, body forces and the viscous forces. This equation is based on Newton’s Second law of motion which holds that acceleration is directly proportional to the force exerted and the force acts in the direction of acceleration. (Wesseling, 2000) With regard to this, it is therefore important to note that when solving this equation, both the prevailing inlet and outlet boundary conditions must be put into consideration. This ensures that these can be used to compute the acceleration as well as the momentum. 3. Conservation of energy This equation states that the temperature rise of a flame coupled with the convective heat transfer is equal to the change in pressure over the same time plus heat released per unit volume from reaction less energy transferred to evaporating droplets and diffusion energy in addition to any other heat source3. The basis for this equation is the first law of thermodynamics which is itself an application of the principle of conservation of energy for thermodynamic and heat systems. The law holds that energy can neither be created nor destroyed. Thus, the energy flowing within a CFD is maintained throughout the simulation. All the boundary conditions therefore need to be considered when solving CFD problems using this equation so as to link the surroundings and the simulation model and also define the net interchange of energy from the surroundings to the model and vice versa. Some of the typical Boundary Conditions which are usually used in solving the aforementioned equations include: Dirichlet – This is a boundary condition that is enforced on an ordinary or partial differential equation. It specifies the values that a solution to a problem should take on along the limits of the area. In CFD analysis the no slip condition for viscous fluids posits that at a solid boundary a fluid will have zero velocity relative to the limit. Nuemann – This is a boundary condition that when enforced on an ordinary or a partial differential equation, stipulates the values that the derivative of a solution is to take on the limits of the area affected. Robin – This is a boundary condition which when enforced on an ordinary or a partial differential equation, stipulates a linear combination of the values of a function and the values of its derivative on the limits of an area in consideration. b) CFD codes are written in low speed solvers when the flow conditions being considered comprise of incompressible elements and hence the simulation majorly relies on the use of a pressure field for computations and simulations. In this case, the pressure field is determined by manipulating the equation of state, and the momentum as well as continuity equations. Conversely, the CFD codes are written in high speed solvers when the flow is comprised elements moving at high velocities and are compressible. Under this approach, the continuity equation is made use of to determine the density field which is in turn utilized within the code to obtain the simulation. (Anselmet, Gagne, Hopfinger and Antonia, 1984) The student will not be able to obtain credible results due to the fact that the object is moving at a speed that is close to that of the speed of sound. This will force the student to use small steps during the simulation in order to justify all the information regarding the object. Due to this, it will make it extremely difficult to practically simulate an accurate true depiction of these object without making massive errors. c) In FDS, the pressure is disintegrated into a temporally-varying background pressure and a spatially-varying agitation that drives the fluid flow. The background pressure is one of the elements that must be considered when dealing with any CFD with regard to FDS due to the fact that they bear upon the results of computations. Different rooms have different background pressures since a volume (room) within a computational domain is isolated from other rooms and is assigned its own background pressure. (McGrattan, 2000) The pressure within the nth zone is a linear combination of its background component and the flow induced perturbation that is represented by the equation below: P(x, t) = 2. Turbulence Turbulence models are required since they have a sharp deviation in their equivalent magnitudes in space and time. They are quite difficult to get a solution for; so they act as a boundary to avoid a fire simulation to go in that particular direction (Cebeci,2004). Turbulence flow is composed of many different features. Therefore it is important for a CFD model to be able to predict as many of them as possible. . This is due to the fact that the nature with which a fluid flows changes at every instant and therefore generalizing it could spell disaster for any problem solving. To get the true picture at every moment that the fluid is under turbulent flow, it is of key importance to model all these elements. LES is a technique used in CFD mostly to simulate the models dissipative forces. These forces vary in nature and include forces such as viscosity- which is the friction between overlying fluid layers,, material diffusivity that occur at length scales smaller than those that are resolved on the numerical grid. These spatial resolutions make it easier for analysts and specialists to investigate their properties, nature and magnitudes thus enabling them to come up with well- informed conclusions to any given CFD problem. The parameters cannot be used directly in the practical simulations of this method. DNS is a method that models on the numerical grid on the order of 1mm and below. It is also possible to use molecular properties μ, k and D. DNS is one of the most useful tools in the field of CFD since it enable the modelling of simulations that would otherwise present a lot of problems in terms of scales if other techniques were to be made use of due to the small scales involved. RANS are equations of modelling fluid flow which are time averaged. It involves a process whereby an immediate magnitude is disintegrated into its time averaged and unstable quantities. This makes it possible to model the flow throughout the simulation and at each and every instant thus enabling analysis at every point. With this at hand, accurate computations can be done with regard to the specific interval of time or instant in time that is being considered. The parameters of the LES model include: Viscosity; provides a stabilizing effect in the numerical algorithm and damping out numerical instabilities that arise in the floe field. It is also important since it gives a picture of the shear forces that exists between the different layers of the fluid and also the fact that it has a bearing on the speed with which the fluid flows. Furthermore, the parameter has a huge bearing on the turbulence experienced in the fluid. Thermal conductivity; the ability of a material to conduct heat. The viscosity of any fluid is affected by temperature and consequently heat. It is therefore one of the parameters that cannot be ignored in coming up with a simulation of the FDS model for any given material. Material diffusivity; this is the ability of a material to conduct thermal energy relative to its ability to store thermal energy. Determination of the flow and storage of thermal energy is one of the reasons why CFD is done. Therefore, determining the diffusivity of a given material is a key feature. Dissipation rate; this is the rate at which kinetic energy is converted to thermal energy by viscosity. The effect of kinetic energy on materials is profound. Therefore, to determine how much of this that is generated within a material is a key objective and hence one of the parameters that must be put into consideration. b) Kolmogorov’s theory states that the energy density per unit wave number should depend only upon the wave number, the rate of energy dissipation per unit volume. The upper limit of the inertial range should depend only on the molecular viscosity. It describes how energy is transfer from larger to smaller eddy; how much energy is contained by eddies of a given size; and how much energy is dissipated by eddies of each size. The mesh resolution required by DNS is a structured one, while on the other hand LES requires complicated modelling. Structured mesh can only be used in box domain topologically. A structured mesh can be transferred to unstructured mesh but the reverse is not true. Unstructured mesh can model much complex geometry but is still limited on some components. The algorithm for generating unstructured mesh is more complex than that for structured mesh.(Pope, 2000) 3. Combustion The simulation of fires in CFD is required in order to resolve the length scales that range from the characteristic of the combustion process to those characteristic of the mass and energy transport throughout an entire compartment. It is possible to create a combustion model which can track important species necessary to calculate heat release rate; consequently it is too costly to create a grid, fine enough to resolve individual flame sheets except in cases where the field is very small. A simulation is thus needed to model the combustion reaction in a method that can be used at length scales of the resolvable flow field. (McGrattan, 2000) A mixture fraction assumes that the combustion reaction is takes place on an infinitely thin flame sheet where both the fuel and the oxygen concentrations go to zero. Mixture fraction also assumes that fuel and oxygen cannot co-exist. It assumes that oxygen and fuel will react at any temperature. FDS can be used to simulate any pre-mixed combustion and explosion, but the model parallels an explosive combustion and features significant pressure deviations. a. Mixture fraction Z is defined as Where - mass fraction of fuel - mass fraction of oxygen - fuel mass fraction in the fuel stream - mass fraction of oxygen at infinity - ratio of oxygen and fuel molecular weight in stoichiometric mixture It is known that =0.94, =0.23 and s = 0.64. Calculate Z on the flame front (5 marks). If at some position = 0.78, calculate Z at this position (5 marks). When use a value that is not given, please explain the reason for taking that value. a) S = ratio of oxygen and fuel molecular weight in stoichiometric mixture = mass fraction of oxygen Mass fraction of fuel (McGrattan and Forney, 2000) = Y0 / YF = 0.64 = 64/100 = 16 / 25 = 16: 25 Therefore Y0 = 16 YF = 2 Z = (0.64 * 25) – (16 – 0.23) (0.64 * 0.94) + 0.23 = 0.23 / 0.8316 = 0.2765 If YF = 0.78 Then Y0 = s * YF Yo = 0.64 * 0.78 = 0.4992 Z = (0.64 * 0.78) – (0.4992 – 0.23) (0.64 * 0.94) + 0.23 Z = 0.23 / 0.8316 = 0.2765 In the first part of the problem Y0 and YF were not provided. We had to get the two values using the identity equation provided that is; - Ratio of oxygen and fuel molecular weight in stoichiometric mixture We had to determine the ratios of the oxygen and fuel mass mixtures utilizing the value of the s given as 0.64. Through simple mathematical relationships we figured that the ratio of oxygen mass to fuel mass is 16: 25 4. Numerical Techniques For the 1D problem the equation can be reduced to: (ΔT/Δt) – (kΔ2T/Δx2) = 0 ……………………………… (1) temp flow(Oran and Boris, 1987) For any case: (ΔT/Δt) ≈ (Til+1 – Til) / (Δt) …………………………… (2) For explicit scheme: (Δ2T/Δx2) = (Ti+1l – 2Til + Ti-1l) / (Δx) 2 ………………. (3) Substitute (2) into (3) to yield (Til+1 – Til) / (Δt) – k (Ti+1l – 2Til + Ti-1l) / (Δx) 2 = 0 Let ƛ = k * Δt / (Δx) 2, then the above equation can be rearranged to: Til+1 – Til – ƛ (Ti+1l – 2Til + Ti-1l) = 0 Or Til+1 = Til + ƛ (Ti+1l – 2Til + Ti-1l) …………. (4) At time t =0 T00 = 22 Ti0 = 0 (i≠0) After one time step at t = Δt, equation 4 can be written as for each node. T01 = 22 T11 = T01 + ƛ (T20 – 2T10 + T00) T21 = T20 + ƛ (T30 – 2T20 + T10) T31 = T30 + ƛ (T40 – 2T30 + T20) T41 = T40 + ƛ (T50 – 2T40 + T30) T51 = 0 Solving the above equations will give T11 values At next time step t = 2Δt Equation 4 can write as: T02 = 22 T12 = T11 + ƛ (T21 – 2T11 + T01) T22 = T21 + ƛ (T31 – 2T21 + T11) T32 = T31 + ƛ (T41 – 2T31 + T21) T42 = T41 + ƛ (T51 – 2T41 + T31)……………………………. (6) T52 = 0 Solving this equation will give T02, T22, T32, T42, T52 ………………. (7) Repeat 6 and 7 until t =15 which give the values T010, T110, T210, T310, T410, T510 Therefore the temperatures at t = 1s will be 22, T110, T210, T310, T410, T510, 0 For implicit scheme replace equation (3) by When a Dirichlet boundary condition is imposed at the two ends, we get; Thus giving us the following system of equations. , i = 1, 2,…, n + Question 4b solution Boundary conditions 240C 00C Temp flow 0.11m ∆t= 0.1 sec ∆x=0.022 No of nodes=+1 =6 nodes Calculation of µ µ= k* (Anderson, Tannehil and Pletcher, 1984) = (0.835*10-4 m2 /s) * =0.0173 At t=0.1 sec T01= 24 T11= T10+µ (T20-2T10+T00) =0+0.0173 (0-2*0+24) =9.9648 T21= T20+µ (T30-2T20+T10)10 =0+0.0173(0-2*0+0) =0.0173 T31= T30+µ (T40-2T30+T20) =0+0.0173(0-2*0+0) =0.0173 T41= T40+µ (T50-2T40+T30) =0+0.0173(0-2*0+0) =0.0173 T51= 0 At t=0.2 sec T02= 24 T11= T11+µ (T21-2T11+T01) =9.9648+0.0173(0.0173-2*9.9648+24) =10.0355 T22= T21+µ (T31-2T21+T11) =0.0173+0.0173(0.0173-2*0.0173+9.9648) =0.1840 T32= T31+µ (T41-2T31+T21) =0.0173+0.0173(0.0173-2*0.0173+0.0173) =0.0173 T42= T41+µ (T51-2T41+T31) =0.0173+0.0173(0-2*0.0173+0.0173) =0.0170 T52= 0 At t=0.3 sec] T03= 24 T13= T12+µ (T22-2T11+T01) =10.035+0.0173(0.1840-2*10.0355+24) =10.067 T23= T22+µ (T32-2T21+T11) =0.1840+0.0173(0.0173-2*0.1840+10.0355) =0.3515 T33= T32+µ (T42-2T31+T21) =0.0173+0.0173(0.0170-2*0.0173+0.1840) =0.0202 T43= T42+µ (T52-2T41+T31) =0.0170+0.0173(0-2*0.0170+0.0173) =0.0167 T53= 0 At t=0.4 sec T04= 24 T14= T13+µ (T21-2T11+T01) =10.1067+0.0173(0.3515-2*10.1067+24) =10.1783 T24= T23+µ (T31-2T21+T11) =0.3515+0.0173(0.0202-2*0.3515+10.1067) 0.5145 T34= T33+µ (T41-2T31+T21) =0.0202+0.0173(0.0167-2*0.0202+0.3515) =0.0259 T44= T43+µ (T51-2T41+T31) =0.0167+0.0173(0-2*0.0167+0.0202) =0.0165 T54= 0 At t=0.5 sec T05= 24 T15= T14+µ (T24-2T14+T04) =10.1783+0.0173(0.5145-2*10.1783+24) =10.2502 T25= T24+µ (T34-2T24+T14) =0.5145+0.0173(0.0259-2*0.5145+10.1783) =0.6732 T35= T34+µ (T44-2T34+T24) =0.0259+0.0173(0.0165-2*0.0259+0.5145) =0.0342 T45= T44+µ (T54-2T44+T34) =0.0165+0.0173(0-2*0165+0.0259) =0.0164 T55= 0 At t=0.6 sec T06= 24 T16= T15+µ (T25-2T15+T05) =10.2502+0.0173(0.6732-2*10.2502+24) =10.3224 T26= T25+µ (T35-2T25+T15) =0.6732+0.0173(0.0342-2*0.6732+10.2502) =0.8278 T36= T35+µ (T45-2T35+T25) =0.0342+0.0173(0.0164-2*0.0342+0.6732) =0.0449 T46= T45+µ (T55-2T45+T35) =0.0164+0.0173(0-2*0.0164+0.0342) =0.0164 T56= 0 At t=0.7 sec T07= 24 T17= T16+µ (T26-2T16+T06) =10.3224+0.0173(0.8278-2*10.3224+24) =10.3948 T27 = T26+µ (T36-2T26+T16) =0.8278+0.0173(0.0449-2*0.8278+10.3224) =0.9785 T37= T36+µ (T46-2T36+T26) =0.0449+0.0173(0.0164-2*0.0449+0.8278) =0.0580 T47= T46+µ (T56-2T46+T36) =0.0164+0.0173(0-2*0.0164+0.0449) =0.0166 T57= 0 At t=0.8 sec T08= 24 T18= T17+µ (T27-2T17+T07) =10.3948+0.0173(0.9785-2*10.3948+24) =10.4673 T28= T27+µ (T37-2T27+T17) =0.9785+0.0173(0.0580-2*0.9785+10.3948) =1.1255 T38= T37+µ (T47-2T37+T27) =0.0580+0.0173(0.0166-2*0580+0.9785) =0.0732 T48= T47+µ (T57-2T47+T37) =0.0166+0.0173(0-2*0.0166+0.0580) =0.0170 T58= 0 At t=0.9 sec T09= 24 T19= T18+µ (T28-2T18+T08) =10.4673+0.0173(1.1255-2*10.4673+24) =10.5398 T29= T28+µ (T38-2T28+T18) =1.1255+0.0173(0.0732-2*1.1255+10.4673) =1.2689 T39= T38+µ (T48-2T38+T28) =0.0732+0.0173(0.0170-2*0.0732+1.1255) =0.0904 T49= T48+µ (T58-2T48+T38) =0.0170+0.0173(0-2*0.0170+0.0732) =0.0177 T59= 0 At t=1 sec T010= 24 T110= T19+µ (T29-2T19+T09) =10.5398+0.0713(1.2689-2*10.5398+24) =10.6123 T210= T29+µ (T39-2T29+T19) =1.2689+0.0173(0.0904-2*1.2689+10.5398) =1.4089 T310= T39+µ (T49-2T39+T29) =0.0904+0.0173(0.0177-2*0.0904+1.2689) =0.1095 T410= T49+µ (T59-2T49+T39) =0.0177+0.0173(0-2*0.0177+0.0904) =0.0187 T510= 0 Therefore the temperature distribution on the thin insulated rod at t=1 sec will be: T0=240C T1=10.61230C T2=1.40890C T3=0.10950C T4=0.01870C T5=00C References: 1. J H Ferziger and M Peric, “Computational Methods for Fluid Dynamics”, Springer Verlag, 1999, 2nd edition 2. P Wesseling, “Principles of Computational Fluid Dynamics”, Springer Verlag, 2000 3. Anselmet, F., Gagne, Y., Hopfinger, E. & Antonia, R. A. 1984 High-order velocity structure functions in turbulent shear flows. J. Fluid Mech. 140, 63-89. 4. Pope, Stephen B. “Turbulent Flows.” Cambridge University Press 2000. 5. McGrattan, K., et al., “Fire Dynamics Simulator - Technical Reference Guide”, National Institute of Standards and Technology, NISTIR 6467, 2000. 6. McGrattan, K. And Forney, G., “Fire Dynamics Simulator - User’s Manual”, National Institute of Standards and Technology, NISTIR 6469, 2000. 7. E.S. Oran and J.P. Boris. Numerical Simulation of Reactive Flow. Elsevier Science Publishing Company, New York, 1987. 8. D.A. Anderson, J.C. Tannehill, and R.H. Pletcher. Computational Fluid Mechanics and Heat Transfer. Hemisphere Publishing Corporation, Philadelphia, Pennsylvania, 1984. 9. CEBECI, T. (2004). Turbulence models and their application: efficient numerical methods with computer programs; with a CD-ROM. Berlin [u.a.], Springer [u.a.]. Read More